Question

$N_1$ and $N_2$ are the magnitudes of normal reactions in between the blocks kept on a smooth horizontal floor as shown in figures (a) and (b). The ratio $\frac{N_1}{N_2}$ is

• A. 1 : 1
• B. 1 : 2
• C. 2 : 1
• D. 2 : 3

Answer 1

The correct option is B 1 : 2


From figure (a):
External force, F = 10 N
Total mass, m = 10 kg + 5 kg = 15 kg
According to Newton’s second law:
F = ma
$\Rightarrow a=\frac{F}{m}=\frac{10}{15}=\frac{2}{3}m{s}^{-2}$
Let $N_1$ be the reaction force acting on the block of mass 5 kg.

$N_1=5a$
$N_1=5\times \frac{2}{3}=\frac{10}{3}N$.......(i)
From figure (b):
External force, F = 10 N
Total mass, m = 10 kg + 5 kg = 15 kg
According to Newton’s second law:
F = ma'
$\Rightarrow a^{\prime }=\frac{F}{m}=\frac{10}{15}=\frac{2}{3}m{s}^{-2}$
Let $N_2$ be the reaction force acting on the block of mass 10 kg.

$N_2=10a^{\prime }$
$N_2=10\times \frac{2}{3}=\frac{20}{3}N$ .....(ii)
Taking the ratio of $N_1$ and $N_2$ from (i) and (ii), we get:
$\frac{N_1}{N_2}=\frac{\frac{10}{3}}{\frac{20}{3}}=\frac{1}{2}$
$\Rightarrow N_1:N_2=1:2$
Hence, the correct answer is option (2).